Midterm 1 (Practice 2)

• (a) The value of the investment is USD \( 130 \).

• (b) The value of the investment in US dollars is \(V=100\cdot (1+r)^3=100\cdot 1.1^3=133.1\).

The price of the bond is: \[P=\frac{C}{1+r}+\frac{C}{(1+r)^2}+\frac{C}{(1+r)^3}+\frac{F}{(1+r)^3}, \] where \(C=54\), \(F=270\), and \(r=0.5\). Since \(1+r=1.5=\frac32\), the previous formula gives us \(P=156\).

• (a) The expected return of the first security is \( \mu_1=\frac{2\cdot 8-8\cdot 2}{100}=0\% \). Similarly, \( \mu_2=20\% \). The risk of the first security is \( \sigma_1=\sqrt{\mathbb E[K_1^2]-\mu_1^2}=\sqrt{0.16}=0.4 \) and the risk for the second security is \( \sigma_2=\sqrt{0.04}=0.2 \).

• (b) The covariance matrix is \[ C=\left[\begin{array}{cc} \sigma_1^2& \mbox{cov }[K_1,K_2]\newline \mbox{cov }[K_1,K_2]&\sigma_2^2\end{array}\right],\] where \( \mbox{cov }[K_1,K_2]=\mathbb E[K_1K_2]-\mathbb E[K_1]\mathbb E[K_2]=\mathbb E[K_1K_2]= \frac{-800}{10000} \). Thus \[ C=\frac1{10000}\left[\begin{array}{cc} 1600&-800\newline -800&400\end{array}\right] =\left[\begin{array}{cc} 0.16&-0.08\newline -0.08&0.04\end{array} \right] . \]

We will use the formula for the weights of the market portfolio: \[ w_{M}=\frac{(m-Ru)C^{-1}}{(m-Ru)C^{-1}u^T}.\] Using the theorem for the inverse of block matrices whose off-diagonal blocks are \( 0 \) we obtain: \[ C^{-1}=\left[\begin{array}{cccc}5 & -3&0&0\newline -3&2&0&0\newline 0&0&5&-2\newline 0&0&-2&1\end{array}\right].\] We now have \[ (m-Ru)C^{-1}u^T=\frac1{10}\left[\begin{array}{cccc} -1&0&1&-1\end{array}\right]\cdot \left[\begin{array}{cccc}5 & -3&0&0\newline -3&2&0&0\newline 0&0&5&-2\newline 0&0&-2&1\end{array}\right] \cdot \left[\begin{array}{c} 1\newline 1\newline 1\newline 1\end{array}\right]=\frac{2}{10}.\] We therefore get \[ w_{M}=\frac1{2}\left[\begin{array}{cccc} -5&3&7&-3\end{array}\right].\]

Let us denote \(w_1=(10\%, 20\%, 30\%, 40\%)\) and \(w_2=(15\%, 25\%, 35\%, 25\%)\). According to Two-Fund Theorem, any portfolio \(w\) on the minimum variance line can be expressed as \(w=\alpha w_1+(1-\alpha)w_2\) for some real number \(\alpha\). The expected return of such a portfolio is \(\mu_w=\alpha \mu_{w_1}+(1-\alpha)\mu_{w_2}=5\alpha+10(1-\alpha)=10-5\alpha\). If the expected return is \(8\), then we get \(10-5\alpha=8\) which implies that \(\alpha=\frac25\). Then we get \(w=\left(13\%, 23\%, 33\%, 31\%\right)\).

We need to find the maximum of the function \(F\) under the constraint \(G(w)=0\) where \(G(w)=uw^T-1\), where \(u=[\begin{array}{cccc}1&1&\dots& 1\end{array}]\). Notice that we have the representation \(F(w)=\frac{wm^T}{wCw^T}\) and that \begin{eqnarray*} \nabla F(w)&=&\frac{\left(wCw^T\right)m-2\left(wm^T\right)wC}{\left(wCw^T\right)^2}=\frac{\sigma^2(w)m-2\mu(w) wC}{\sigma^4(w)}. \newline \nabla G(w)&=&u. \end{eqnarray*}

We will use the method of Lagrange multipliers to find the critical points of \(F\) under the constraint \(G(w)=0\). There are 5 categories of critical points.

• \(1^{\circ}\) Endpoints. There are no critical points in this category since weights can be any real numbers.

• \(2^{\circ}\) Points for which \(\nabla F\) or \(\nabla G\) is not defined. The gradients \(\nabla F\) and \(\nabla G\) are defined for all weights \(w\).

• \(3^{\circ}\) Points for which \(\nabla F=0\). If \(\nabla F=0\) then we would have \(\sigma^2(w)m-2\mu(w)wC=0\). Multiplying both sides by \(w^T\) from the right implies \(\sigma^2(w)\mu(w)-2\mu(w)\sigma^2(w)=0\) which would imply that \(\mu(w)=0\). For such a critical point we would have \(F(w)=0\), which clearly is not a candidate for maximum.

• \(4^{\circ}\) Points for which \(\nabla G=0\). Since \(\nabla G=u\neq 0\), there are no critical points in this category.

• \(5^{\circ}\) Points for which \(\nabla F\|\nabla G\). We need to find the vectors \(w\) for which \(wu^T=1\) and \(\nabla F(w)=\lambda \nabla G(w)\) for some scalar \(\lambda\). The last equation becomes \[\sigma^2 m-2\mu wC=\lambda \sigma^4u.\quad\quad\quad\quad\quad (1)\] If we multiply both sides of the equation from the right by \(w^T\) we obtain \(\sigma^2 \mu-2\mu\sigma^2=\lambda\sigma^4\) which means that \[\lambda=-\frac{\mu}{\sigma^2}.\] This \(\lambda\) still depends on \(\mu\) and \(\sigma\). We substitute this \(\lambda\) in the equation (1) and multiply both sides of the obtained equation by \(C^{-1}\) from the right. The equation becomes \(\sigma^2mC^{-1}-2\mu w=-\mu\sigma^2 uC^{-1}\). Solving for \(w\) gives us \[w=\frac{\sigma^2}{2\mu}\left(mC^{-1}+\mu uC^{-1}\right).\quad\quad\quad\quad\quad (2)\] We now multiply from the right both sides of (2) by \(u^T\) and \(m^T\). We obtain the following two equations \begin{eqnarray*}1&=&\frac{\sigma^2}{2\mu}\left(mC^{-1}u^T+\mu uC^{-1}u^T\right)\quad\quad\quad\quad\quad\quad (3)\newline \mu&=&\frac{\sigma^2}{2\mu}\left(mC^{-1}m^T+\mu uC^{-1}m^T\right).\quad\quad\quad\quad\quad (4) \end{eqnarray*} If we now multiply the equations (3) by \(\mu\) and subtract it from (4) we get \(\mu^2 uC^{-1}u^T-mC^{-1}m^T=0\). A negative value for \(\mu\) would lead to a negative value for \(F\) which clearly cannot be the maximum. Therefore \(\mu > 0\) and \[\mu=\sqrt{\frac{mC^{-1}m^T}{uC^{-1}u^T}}.\] From the equation (3) we now get \[\frac{\mu}{\sigma^2}=\frac12\left(mC^{-1}u^T+\mu uC^{-1}u^T\right)= \frac12\left(mC^{-1}u^T+\sqrt{mC^{-1}m^T\cdot uC^{-1}u^T}\right).\quad\quad\quad\quad\quad(5)\] We can solve the last equation for \(\sigma\) and obtain \(w\) from (2). However, since the problem only asks for the maximum of the function, and since \(F(w)=\mu/\sigma^2\) we conclude that the value of the function for this particular critical point \(w_0\) is given by (5).

Thus the maximum of the function is \[F_{\max}=\frac12\left(mC^{-1}u^T+\sqrt{mC^{-1}m^T\cdot uC^{-1}u^T}\right).\]